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## Homework Statement

Find the point on the x-axis where the net electric field is zero for two particles of charges q1=1x10^-9C and q2=2x10^-9C. Assume q1 and q2 are 20cm apart. Assume q1 is located at x=0.

I solved this but it's not the same as the book's answer.. GAH! What do you guys get? Thanks.

## Homework Equations

Enet=E1+E2, Enet=0 => |E1|=|E2|

E=kq/r² where q is the charge creating the field

## The Attempt at a Solution

E1=kq1/r²

E2=kq2/(.2m - r)²

kq1/r²=kq2/(.2m - r)²

q1/r²=q2/(.2m - r)²

q1(.2m - r)²=q2r²

q1(.2m - r)²=q2r²

q1(0.4m-0.4mr+r²)-q2r²=0 <------ There's my mistake

(q1)0.4m-(q1)0.4mr+(q1)r²-q2r²=0

(q1-q2)r²-(q1)0.4mr+(q1)0.4m=0

((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.4m=0

(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.4m=0

(-1E-9C)r²-(4E-10C)mr+(4E-10C)m=0

(-1E-9)r²-(4E-10)r+(4E-10)=0

r=-0.863325 or r=0.463325

The answer in the book is 8.3cm. Where did I go wrong?

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Edit:

q1(.2m - r)²=q2r²

q1(0.04m-0.4mr+r²)-q2r²=0

(q1)0.04m-(q1)0.4mr+(q1)r²-q2r²=0

(q1-q2)r²-(q1)0.4mr+(q1)0.04m=0

((1E-9C)-(2E-9C))r²-(1E-9C)0.4mr+(1E-9C)0.04m=0

(-1E-9C)r²-(1E-9C)0.4mr+(1E-9C)0.04m=0

(-1E-9C)r²-(4E-10C)mr+(4E-11C)m=0

(-1E-9)r²-(4E-10)r+(4E-11)=0

Positive r comes out to 8.3cm

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